Rotational Motion Question 26

Question: A solid sphere is rolling on a surface , with a translational velocity $ vm{s^{-1}}. $ If it is to climb the inclined surface continuing to roll without slipping, then minimum velocity for this to happen is

[JEE ONLINE 12-05-2012]

Options:

A) $ \sqrt{2gh} $

B) $ \sqrt{\frac{7}{5}gh} $

C) $ \sqrt{\frac{7}{2}gh} $

D) $ \sqrt{\frac{10}{7}gh} $

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Answer:

Correct Answer: D

Solution:

  • [d] Minimum velocity for a body rolling without slipping For solid sphere,

$\frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 = mgh$

$\frac{1}{2} m v^2 + \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v^2}{r^2}\right) = mgh$

$\frac{v^2}{2} + \frac{2v^2}{10} = g h$

$\frac{5v^2}{2} + \frac{2v^2}{10} = g h \Rightarrow v^2 = \frac{10}{7} gh$

$v \geq \sqrt{\frac{10}{7} gh}$



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